can someone please explain to me how 27^(2/21) becomes 9^(1/7). I'm so confused and the textbook does not have an example like this, only simpler ones. A walkthrough would be greatly appreciated.
2007-03-04 05:50:14 · 6 answers · asked by Harry W 2 in Science & Mathematics ➔ Mathematics
27 = 3^3 (3 cubed).
27^(2/21) = 3^(3*2/21) = 3^(2/7) = (3^2)^(1/7) = 9^(1/7).
Hope that's clear.
2007-03-04 05:59:46 · answer #1 · answered by Anonymous · 0⤊ 0⤋
27^(2/21)=3^3^(2/21) and a^m^n=a^m*n
=3^3*2/21
=3^2(1/7)
=9^(1/7)
2007-03-04 14:02:17 · answer #2 · answered by psk 1 · 0⤊ 0⤋
27 ^ ( 2/21) = (3 ^3) ^ (2/21) = 3 ^ (3 * 2/21) = 3^(2/7) = (3^2)^(1/7) = 9^(1/7)
2007-03-04 13:55:54 · answer #3 · answered by hustolemyname 6 · 0⤊ 0⤋
27 = 3^3, so
27^(2/21) = (3^3)^(2/21) (switch the 3 and 2 by rules of exponents)
= (3^2)^(3/21) = 9^(1/7)
2007-03-04 13:55:38 · answer #4 · answered by Anonymous · 0⤊ 0⤋
27^(2/21)
= (3^3)^(2/21)
= 3^(2/7)
= (3^2)^(1/7)
= 9^(1/7)
2007-03-04 13:57:38 · answer #5 · answered by sahsjing 7 · 0⤊ 0⤋
27^(2/21) = 27^(2/3 * 1/7) = (27^(2/3))^(1/7)
This is the cube root of 27 squared to the one-seventh.
The cube root of 27 is 3, and 3 squared is 9.
Hence, 9^(1/7)
2007-03-04 13:55:15 · answer #6 · answered by Dave 6 · 0⤊ 0⤋